Answer
a.\begin{aligned}
f'(x)&=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{x+h-x}\\
&=\lim\limits_{h\to0}\frac{\sqrt{3-5(x+h)}-\sqrt{3-5x}}{h} \\
&=\lim\limits_{h\to0}\frac{(\sqrt{3-5(x+h)}-\sqrt{3-5x})(\sqrt{3-5(x+h)}+\sqrt{3-5x})}{h(\sqrt{3-5(x+h)}+\sqrt{3-5x})} \\
&=\lim\limits_{h\to0}\frac{3-5(x+h)-3+5x}{h(\sqrt{3-5(x+h)}+\sqrt{3-5x})} \\
&=\lim\limits_{h\to0}\frac{-5h}{h(\sqrt{3-5(x+h)}+\sqrt{3-5x})} \\
&=\frac{-5}{(\sqrt{3-5x}+\sqrt{3-5x})} \\
&=\frac{-5}{2\sqrt{3-5x}}
\end{aligned}
Work Step by Step
b.For $f(x)$,$3-5x\geq 0$,so $x\leq\frac 35$.f's domain is $\{x\vert x\leq\frac 35\}$
For $f'(x)$,$3-5x\geq0$ and $\sqrt{3-5x}\neq 0$
So $x\neq \frac 53$
So $f'$'s domain is $\{x\vert x<\frac35\}$
